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3.Current Electricity
hard
Two resistances are connected in two gaps of a meter bridge. The balance point is $20\, cm$ from the zero end. A resistance of $15\, ohms$ is connected in series with the smaller of the two. The null point shifts to $40\, cm$. The value of the smaller resistance in $ohms$ is
A
$3$
B
$6$
C
$9$
D
$12$
Solution
(c) Let $S$ be larger and $R$ be smaller resistance connected in two gaps of meter bridge.
$S = \left( {\frac{{100 – l}}{l}} \right)R = \frac{{100 – 20}}{{20}}R = 4R$ …..$(i)$
When $15\,\Omega $ resistance is added to resistance $R$, then
$S = \left( {\frac{{100 – 40}}{{40}}} \right)(R + 15) = \frac{6}{4}(R + 15)$ …. $(ii)$
From equations $(i)$ and $(ii)$ $R = 9\,\Omega $
Standard 12
Physics
